Integrand size = 17, antiderivative size = 91 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-a B+A b x}{7 a b \left (a+b x^2\right )^{7/2}}+\frac {6 A x}{35 a^2 \left (a+b x^2\right )^{5/2}}+\frac {8 A x}{35 a^3 \left (a+b x^2\right )^{3/2}}+\frac {16 A x}{35 a^4 \sqrt {a+b x^2}} \] Output:
1/7*(A*b*x-B*a)/a/b/(b*x^2+a)^(7/2)+6/35*A*x/a^2/(b*x^2+a)^(5/2)+8/35*A*x/ a^3/(b*x^2+a)^(3/2)+16/35*A*x/a^4/(b*x^2+a)^(1/2)
Time = 0.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-5 a^4 B+35 a^3 A b x+70 a^2 A b^2 x^3+56 a A b^3 x^5+16 A b^4 x^7}{35 a^4 b \left (a+b x^2\right )^{7/2}} \] Input:
Integrate[(A + B*x)/(a + b*x^2)^(9/2),x]
Output:
(-5*a^4*B + 35*a^3*A*b*x + 70*a^2*A*b^2*x^3 + 56*a*A*b^3*x^5 + 16*A*b^4*x^ 7)/(35*a^4*b*(a + b*x^2)^(7/2))
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {454, 209, 209, 208}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 454 |
\(\displaystyle \frac {6 A \int \frac {1}{\left (b x^2+a\right )^{7/2}}dx}{7 a}-\frac {a B-A b x}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {6 A \left (\frac {4 \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a}-\frac {a B-A b x}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {6 A \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a}-\frac {a B-A b x}{7 a b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {6 A \left (\frac {4 \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a}-\frac {a B-A b x}{7 a b \left (a+b x^2\right )^{7/2}}\) |
Input:
Int[(A + B*x)/(a + b*x^2)^(9/2),x]
Output:
-1/7*(a*B - A*b*x)/(a*b*(a + b*x^2)^(7/2)) + (6*A*(x/(5*a*(a + b*x^2)^(5/2 )) + (4*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])))/(5*a) ))/(7*a)
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a *(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L tQ[p, -1] && NeQ[p, -3/2]
Time = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70
method | result | size |
gosper | \(\frac {16 A \,b^{4} x^{7}+56 A a \,b^{3} x^{5}+70 A \,a^{2} b^{2} x^{3}+35 A \,a^{3} b x -5 B \,a^{4}}{35 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b}\) | \(64\) |
trager | \(\frac {16 A \,b^{4} x^{7}+56 A a \,b^{3} x^{5}+70 A \,a^{2} b^{2} x^{3}+35 A \,a^{3} b x -5 B \,a^{4}}{35 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b}\) | \(64\) |
orering | \(\frac {16 A \,b^{4} x^{7}+56 A a \,b^{3} x^{5}+70 A \,a^{2} b^{2} x^{3}+35 A \,a^{3} b x -5 B \,a^{4}}{35 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b}\) | \(64\) |
default | \(A \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )-\frac {B}{7 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\) | \(92\) |
Input:
int((B*x+A)/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/35*(16*A*b^4*x^7+56*A*a*b^3*x^5+70*A*a^2*b^2*x^3+35*A*a^3*b*x-5*B*a^4)/( b*x^2+a)^(7/2)/a^4/b
Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (16 \, A b^{4} x^{7} + 56 \, A a b^{3} x^{5} + 70 \, A a^{2} b^{2} x^{3} + 35 \, A a^{3} b x - 5 \, B a^{4}\right )} \sqrt {b x^{2} + a}}{35 \, {\left (a^{4} b^{5} x^{8} + 4 \, a^{5} b^{4} x^{6} + 6 \, a^{6} b^{3} x^{4} + 4 \, a^{7} b^{2} x^{2} + a^{8} b\right )}} \] Input:
integrate((B*x+A)/(b*x^2+a)^(9/2),x, algorithm="fricas")
Output:
1/35*(16*A*b^4*x^7 + 56*A*a*b^3*x^5 + 70*A*a^2*b^2*x^3 + 35*A*a^3*b*x - 5* B*a^4)*sqrt(b*x^2 + a)/(a^4*b^5*x^8 + 4*a^5*b^4*x^6 + 6*a^6*b^3*x^4 + 4*a^ 7*b^2*x^2 + a^8*b)
Leaf count of result is larger than twice the leaf count of optimal. 1266 vs. \(2 (85) = 170\).
Time = 11.76 (sec) , antiderivative size = 1360, normalized size of antiderivative = 14.95 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\text {Too large to display} \] Input:
integrate((B*x+A)/(b*x**2+a)**(9/2),x)
Output:
A*(35*a**14*x/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt (1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2 )*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/a ) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12* sqrt(1 + b*x**2/a)) + 175*a**13*b*x**3/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b* *4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 371*a**12*b**2*x**5/(35*a** (37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525* a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b** 5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 429*a**11*b**3*x**7/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x* *2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a **(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b *x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6 *x**12*sqrt(1 + b*x**2/a)) + 286*a**10*b**4*x**9/(35*a**(37/2)*sqrt(1 + b* x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x** 4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525...
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {16 \, A x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {A x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {B}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} \] Input:
integrate((B*x+A)/(b*x^2+a)^(9/2),x, algorithm="maxima")
Output:
16/35*A*x/(sqrt(b*x^2 + a)*a^4) + 8/35*A*x/((b*x^2 + a)^(3/2)*a^3) + 6/35* A*x/((b*x^2 + a)^(5/2)*a^2) + 1/7*A*x/((b*x^2 + a)^(7/2)*a) - 1/7*B/((b*x^ 2 + a)^(7/2)*b)
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (2 \, {\left (4 \, {\left (\frac {2 \, A b^{3} x^{2}}{a^{4}} + \frac {7 \, A b^{2}}{a^{3}}\right )} x^{2} + \frac {35 \, A b}{a^{2}}\right )} x^{2} + \frac {35 \, A}{a}\right )} x - \frac {5 \, B}{b}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:
integrate((B*x+A)/(b*x^2+a)^(9/2),x, algorithm="giac")
Output:
1/35*((2*(4*(2*A*b^3*x^2/a^4 + 7*A*b^2/a^3)*x^2 + 35*A*b/a^2)*x^2 + 35*A/a )*x - 5*B/b)/(b*x^2 + a)^(7/2)
Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {16\,A\,x}{35\,a^4\,\sqrt {b\,x^2+a}}-\frac {\frac {B}{7\,b}-\frac {A\,x}{7\,a}}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {8\,A\,x}{35\,a^3\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {6\,A\,x}{35\,a^2\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:
int((A + B*x)/(a + b*x^2)^(9/2),x)
Output:
(16*A*x)/(35*a^4*(a + b*x^2)^(1/2)) - (B/(7*b) - (A*x)/(7*a))/(a + b*x^2)^ (7/2) + (8*A*x)/(35*a^3*(a + b*x^2)^(3/2)) + (6*A*x)/(35*a^2*(a + b*x^2)^( 5/2))
Time = 5.26 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.02 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {35 \sqrt {b \,x^{2}+a}\, a^{3} b x -5 \sqrt {b \,x^{2}+a}\, a^{3} b +70 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{3}+56 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{5}+16 \sqrt {b \,x^{2}+a}\, b^{4} x^{7}-16 \sqrt {b}\, a^{4}-64 \sqrt {b}\, a^{3} b \,x^{2}-96 \sqrt {b}\, a^{2} b^{2} x^{4}-64 \sqrt {b}\, a \,b^{3} x^{6}-16 \sqrt {b}\, b^{4} x^{8}}{35 a^{3} b \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int((B*x+A)/(b*x^2+a)^(9/2),x)
Output:
(35*sqrt(a + b*x**2)*a**3*b*x - 5*sqrt(a + b*x**2)*a**3*b + 70*sqrt(a + b* x**2)*a**2*b**2*x**3 + 56*sqrt(a + b*x**2)*a*b**3*x**5 + 16*sqrt(a + b*x** 2)*b**4*x**7 - 16*sqrt(b)*a**4 - 64*sqrt(b)*a**3*b*x**2 - 96*sqrt(b)*a**2* b**2*x**4 - 64*sqrt(b)*a*b**3*x**6 - 16*sqrt(b)*b**4*x**8)/(35*a**3*b*(a** 4 + 4*a**3*b*x**2 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b**4*x**8))